 5:21 AM
0  Let $f: [0,1] \to \mathbb R$ be a continuous function then the maximum value of $\displaystyle \int_0 ^1 f(x)(x^2) dx - \int_0^1 x(f(x)^2) dx$ is? Unable to understand how to solve this problem. I have thought about it analytically and think that the following points are important: \$x^...

There was a discussion about such tag before.
18  For people who only want guidance in understanding how to approach the problem, as opposed to seeing a fully fleshed out solution. Would this tag be useful? Am new to this site, so don't want to create this tag especially if people think that it won't be used. I don't really use tags as yet, s...

I have removed the tag from this question. Having a separate tag for hints is a big enough change so that it warrants discussion on meta first before creating the tag.
Another new tag is .
0  How to calculate the normalisation factors of Legendre Polynomial of second kind? It is provided that ,the normalisation factors are chosen so that second kind Polynomials satisfies the recurrence relation of the first kind.

Is there a difference between and ? Legendre polynomials
In mathematics, Legendre functions are solutions to Legendre's differential equation: They are named after Adrien-Marie Legendre. This ordinary differential equation is frequently encountered in physics and other technical fields. In particular, it occurs when solving Laplace's equation (and related partial differential equations) in spherical coordinates. The Legendre differential equation may be solved using the standard power series method. The equation has regular singular points at x = ±1 so, in general, a series solution about the origin will only converge for |x| < 1. When n is an integer...

13 hours later… 5:59 PM
0  Suggestion: Rename tag box-product to box-topology. The tag is of low usage (fewer than a dozen instances, some of which are inappropriate if the proper application is for topology. I noticed this tag in reviewing a tag wiki excerpt which presumed that was the intention, but I rejected it as be...